You may have noticed that
always using the midpoint of the interval generated some pretty long numbers.
My computations weren't too hard, because I was making intelligent use
of my calculator. But
there were times when it was pretty obvious that I could have used better,
and easier, x-values,
and still gotten the same answer. If you're working with a calculator
that doesn't allow for the symbolic calculations of the polynomial for
a given value of x,
then choosing x-values
with fewer decimal places could be especially handy. So let's find
the polynomial's zero again, but this time we'll go to the trouble
of picking our the x-values,
instead of just mindlessly always using the midpoint.

Okay, now I'm too high
again, but I'm getting pretty close. Lemme edge down a bit:

x =
1.2994: y = 0.000129009...x
= 1.2993: y = –0.001502678...

Finally! I've managed to
bracket the zero between two x-values
that have the same first three decimal places. That's all the accuracy
I need, so I'm done.

x
= 1.299

As you can see, by picking
your own x-values,
you can simplify and shorten your computations. But you do have to pay
attention to what you're doing, or you could end up picking lots of fairly
useless values and taking nearly as long as you would by using the midpoint
process. Either way, though, numerically approximating zeroes isn't terribly
complicated; it's just long and tedious.

If you don't have a graphing
calculator, just a regular calculator, you may have to do a little more
preliminary work on your own (if they don't tell you "look between
here and here for the zero"), but the general process will still
be the same.

The first thing I have
to do is figure out the general area of the zero. Without a graph and
without limits being given to me, I'll have to use what
I know of polynomials
to judge the values I get. I'll try a few values in the vicinity of
the origin:

x =
0: y = –9.1(I'm
below the axis)x
= 1: y = –8.8(The
graph is coming back up toward the axis)x
= 3: y = –10(The
graph is heading back down; I may be moving the wrong way)x
= 5: y = –137.6(Oh,
yeah, this has got to be the wrong direction; I'll head negative)x
= –1:
y = –9.2(This
is about the same as where I started)x
= –2:
y = –2.5(The
graph is heading back up toward the axis)x
= –3:
y = 36(The
polynomial is finally positive)

Bingo! I've found a sign
change, so the zero is somewhere between x
= –3
and x
= –2.
Also, the polynomial is above the axis at the lower x-value,
so I know the graph goes down as it crosses the axis. Then a negative
y-value
will mean that my x-value
was too high, and a positive y-value
will mean that my x-value
was too low. This will tell me in which direction I need to move.

Since y
= –2.5 (at
x
= –2)
is a lot closer to zero than is y
= 36 (at x
= –3),
the zero must be a lot closer to x
= –2,
so I'll test some values in the tenths:

The sign change tells me
that the zero is bracketed by –2.14 and
–2.13,
with the polynomial being closer to the axis at –2.13.
I'll test values in the thousandths:

x =
–2.133:
y = –0.02469719...(I've
already crossed the axis, so I need to back up)x
= –2.134:
y = –0.003837470...(Still
negative, so I need to back up a little more)x
= –2.135:
y = 0.0170576798...(The
polynomial is back above the axis)

Now I have the zero bracketed
between –2.135 and
–2.134,
numbers that have the same first two decimal places, so I'm done.

x
= –2.13

If you're writing a computer
program or graphing-calculator
program to find
numerical approximations, use the midpoint process: it's mindless and
rote, which is what you need for an automated process like a program.
If your text specifies the method you should use, you should stick with
that. Otherwise, use whichever method you prefer. Just allow yourself
some time, and be sure to write down your steps carefully. About the only
problem you can really have with this process would come from copying
down a number wrong. As long as you're neat and don't rush, you should
do fine.